short circuit analysis lecture notes

SHORT CIRCUIT ANALYSIS

Topics 1. Introduction2. Cause and Consequences3. Types of Faults4. Symmetrical Short Circuit Analysis

SHORT CIRCUIT ANALYSIS

Objective

Explain the significance of Short Circuit Explain the causes and consequences of

Short Circuit

Classify different types of faults Compute currents for symmetrical faults

Cause of Short Circuit Insulation Failure Over-voltages caused by Lightning or

Switching Surges

Insulation contamination - salt spray, pollution Mechanical Causes - Over-heating, abrasion

Faults on Transmission LinesMost Common Lines are exposed to elements of nature (60-70%).

Lightning strokes Over voltages cause insulators to flash over line to ground short circuit or line to line short circuit.

High winds Topple tower, tree falls on line.Winds and ice loading Mechanical failure of insulator.

Fog, salt spray, dirty insulator Conduction path insulation failure

Short circuit in other elements Cables (10-15%), circuit breakers (10-12%), generators, motors, transformers etc (10-15%). much less common Over loading for extended periods deterioration of insulation Mechanical failure.

Consequences of Short Circuit Currents several magnitude larger than

normal operating current.

Thermal damage to equipment. Windings and busbars Mechanical

damage due to high magnetic forces caused by high current.

Faulted section must be removed from service as soon as possible (3-5 cycles).

Types of Short Circuit

abc

L G L L L L G 3 G75 - 80% 5 7% 10 12% 8 10%

Asymmetrical Faults

Symmetrical Faults

Short Circuit CalculationsR L Series Circuit Transients

SW t = 0

R L

e(t)

max

e(t) = 2V Sin ( t + ) = V Sin ( t + )

max diV sin t + = Ri + L ; t 0dt

V Rt/Lmaxi(t) = sin t + - e sin - Z Solving for i(t)

22 -1ac dc

ac

dc

Where, Z = R L and = tan L /R .or i(t) = i (t) + i (t) i (t) = Symmetrical fault current (Constant) i (t) = DC offset current (decays with time) i(t) = Asymmetrical faul

t current

Time

I(t)

idc iac

3-phase Short Circuit on Synchronous Machine

Unloaded Machine:

Time

Actual envelope

Subtransient period Transient period

Steady state period

0

a

bc

S

y

m

m

e

t

r

i

c

a

l

s

h

o

r

t

c

i

r

c

u

i

t

c

u

r

r

e

n

t

XI

Eg

+

Xa

XdSynchronous reactance

XI

Eg

+ Xa

Xf

Direct axis subtransient reactance

XI

Eg

+Xa

Xf

Direct axis transient reactance

XD

'dX

'' 'd d

ac g

'd d d

''d

'd

- t / T

- t / T

1 1- X X

i (t) = 2 E1 1 1+ - +

X X X x Sin ( t + - )2

e

e

A A- t / T - t / Tg ' 'dcmax ''

d

A

2E i (t) = e = 2 I e ;

X T = Armature Time Const.

g ''ac ''

d

EI (0) = OC = = I

X

(Subtransient Current)

g''d

EI = Ob = (Transient current)

X g

acd

EI ( ) = Oa = (Steady State current)

X

ac dci (t) = i (t) + i (t)

Short Circuit on a loaded Synchronous Machine

'' '' ''g t dg L f ext dg LE V jX I V Z jX I ' ' 'g t dg L f ext dg LE V jX I V Z jX I

IL

Zext

Vt Vf

P

ZL+

+ +

S''gE

''dgX

m t dm LE" V jX" I m t dm LE' V jX ' I

For Motor:

Example: A synchronous generator and asynchronous motor each rated 50 MVA, 11KVhaving 12% subtransient reactance areconnected through transformers and a line asshown in figure below. The transformers arerated 50 MVA, 11/132 KV and 132/11KV withleakage reactance of 8% each. The line has areactance of 15% on a base of 50 MVA, 132 KV.

The motor is drawing 25 MW at 0.8 power factor leading and a terminal voltage of 10.6 KV when a symmetrical three-phase fault occurs at the motor terminals. Find the sub-transient current in the generator, motor and fault.

Gen T1 Line T2 Motor

Zext

jXdg

Eg

V0

I0P

+ +

+

(a) Before the faultNeutral

jXdm

Em

(j0.12)

(j0.31)

(j0.12)

Zext

jXdg

Eg

Ig

P

+ +

(b) After the faultNeutral

jXdm

Em

(j0.12)(j0.31)

(j0.12)Im

If

0 010.6V = = 0.9636 0 pu11

= 25 MW 0.8 pf leading25= pu 0.8 pf leading50

= 0.5 pu 0.8 pf leading

Prefault Voltage =

Load

Prefault Current I0= o o0.5 36.9 = 0.6486 36.90.96360.8

Voltage behind sub-transient reactance (generator)

" o ogE = 0.9636 0 + j0.430.6486 36.9 = 0.7962 + j0.223pu

Voltage behind sub-transient reactance (Motor) " o omE = 0.9636 0 - j0.120.6486 36.9

= 1.0103- j0.0622pu

Under faulted condition"g

"m

0.7962 + j0.223I = = 0.5186 - j1.8516puj0.43

1.0103- j0.0622I = = -0.5183 - j8.4191puj0.12

Current in fault f " "g mI = I +I = 0.0003- j10.2707pu

Base current (generator/motor) =35010 = 2624.3A

3 11

"g

"m

f

I =2624.3(0.5186- j1.8516) =(1360.96- j4859.15) A

I =2624.3(-0.5183- j8.4191) =(-1360.17- j22094.24) A

I =2624.3(0.0003- j10.2707) =(0.78- j26953.39) A

Contribution from Gen. and Motor

Short Circuit Calculations using Thevenins Theorem

A Short Circuit Structural change in network addition of an impedance (ZF = fault impedance, zero for solid short circuit) at the point of fault.

The change in voltage or current resulting from this structural network change can be analyzed using Thevenins theorem.

Thevenins Theorem

The changes in network voltages andcurrents due to the addition of animpedance between two points of a networkare identical with those voltages andcurrents that would be caused by placing anemf, having a magnitude and polarity equalto the pre-fault voltage between the nodes,in series with the impedance all othervoltage sources being zeroed.

Post fault voltages and currents are computed by superimposing these changes on pre-fault voltages and currents.


Example:

Zext

jXdg

Eg

V0

I0P

+ +

+

(a) Before the faultNeutral

jXdm

Em

(j0.12)

(j0.31)

(j0.12)

G

Zext

jXdg

Eg V0

I0P

+ +

+

(b) After the faultNeutral

jXdm

Em

(j0.12)

(j0.31)

(j0.12)

G

V0

+

Zext

jXdg

Eg

I0P

+ +

+

Circuit A (Pre fault Circuit)

Neutral

jXdm

Em

(j0.12)

(j0.31)

(j0.12)

G

V0

Using Superposition:

Zext

jXdg

V0

P +

Circuit B (Thevenins Eq. Circuit)

Neutral

jXdm(j0.12)

(j0.31)

(j0.12)

''gI ''mI

G

0 010.6V = = 0.9636 0 pu11Prefault Voltage =

o 0''g

o 0''m

'' 0 '' og g

'' 0 '' om m

V 0.9636 0I = = - j2.2409j0.31 + j0.12 j0.43

V 0.9636 0I = = - j8.03j0.12 j0.12

I = I + I = 0.6486 36.9 - j2.2409 = 0.5186 - j1.8516I = - I + I = 0.6486 36.9 - j8.03 = - 0.5186 - j8.4194

From the viewpoint of current the two factors that need to be considered in selecting circuit breakers are:

The maximum instantaneous current which the breaker must carry ( withstand ) and

The total current when the breaker contacts open to interrupt the circuit.

The selection of circuit breakers

The maximum momentary current is found by calculating the ac short circuit current using sub-transient impedances of the generators and motors and then multiplying it by 1.6 to take care of the dc off-set current.

The breaker interrupting current depends on the interruption time of the circuit breakers and is obtained by multiplying the sub-transient ac short circuit current by following factors:

Circuit breaker speed Multiplying factor

8 cycle or slower 1.0

5 cycles 1.1

3 cycles 1.2

2 cycles 1.4

For CBs having short circuit MVA greater than 500 MVA the multiplying factors are increased by 0.1

SYMMETRICAL COMPONENT ANALYSIS

SYMMETRICAL COMPONENT ANALYSIS

1. Introduction2. Symmetrical Component Transformation3. Sequence network for PS Components4. Sequence network for Power Systems

Objective

Explain the significance of symmetrical component transformation

Develop sequence network for power system components and networks

Compute current, voltage and power in sequence networks

Types of Short Circuit

abc

L G L L L L G 3 G75 - 80% 5 7% 10 12% 8 10%

Asymmetrical Faults

Symmetrical Faults

Symmercal Component Transformation Introduced by C. L. Fortescue (1918)

Modeling technique for analysis and design ofthree-phase systems

Decouples a balanced three-phase networkinto three simpler networks

For unbalanced three phase networks the threesequence networks are connected only at thepoint of unbalance.

Symmerical Component Transformation A powerful tool for analyzing three phase systems

Reveals complicated phenomena during unbalanced operation in simple terms

Sequence network results have to be superposed to obtain three phase network results

SYMMETRICAL COMPONENT TRANSFORMATION

An unbalanced set of n phasors can be resolvedinto n sets of balanced phasors (symmetricalcomponents). The n phasors of each set ofcomponents are equal in magnitude and anglesbetween adjacent phasors of the set are equal.

Unbalanced phasors of a three phase systemcan be resolved into three balanced system ofphasors positive, negative, and zerosequence

Positive-sequence components, consist of three phasors with equal magnitudes with 120 phase displacement from each other, and same phase sequence as original phasors.

Negative-sequence components, consist of three phasors with equal magnitudes with 120 phase displacement from each other, and opposite phase sequence as original phasors.

Vc1

Vb1

Va1 = V1

Va2 = V2Vb2

Vc2

Zero-sequence components, consist of three phasors with equal magnitudes and zero phase displacement from each other.

Va0Vb0Vc0 = V0

Va

Vb

Vc

Va0 Va1Va2

VaVb1

Vb2

Vb0Vb Vc1Vc2

Vc Vc0

Phase a Phase b Phase c

Va = Va0 + Va1 + Va2

Vb = Vb0 + Vb1 + Vb2

Vc = Vco + Vc1 + Vc2

Va = V0 + V1 + V2Vb = V0 + a2V1 + aV2Vc = V0 + aV1 + a2V2

a 02

P b 12

c 2

V 1 1 1 VV = V = 1 a a V

V 1 a a V

o -1 3a = 1120 = + j2 2

22

1 1 1A = 1 a a

1 a a

0

s 1

2

VV = V

Vand

Where,Vp = A Vs

Vs = A-1 Vp Where,

-1 2

2

1 1 11A = 1 a a 3 1 a a

0 a

21 b

22 c

V 1 1 1 V1V = 1 a a V3V 1 a a V

0 a b c1V = (V + V + V )3

21 a b c

1V = (V +aV +a V )3

22 a b c

1V = (V +a V +aV )3

Ip = A Is

Ia = I0 + I1 + I2Ib = I0 + a2I1 + aI2Ic = I0 + aI1 + a2I2

Is = A-1 Ip

0 a b c1I = (I +I +I )3

21 a b c

1I = (I +aI +a I )3

22 a b c

1I = (I +a I +aI )3

POWER IN SEQUENCE NETWORKS

* * *3 ag a bg b cg cS = V I + V I + V I

T *p p= V I

*a*

3 ag bg cg b*c

IS = V V V I

I

T *3 s sS = (AV ) (AI )

T *

T * 2 2

2 2

1 1 1 1 1 1A A = 1 a a 1 a a

1 a a 1 a a

T T * *s s= V A A I

2 2

2 2

1 1 1 1 1 1= 1 a a 1 a a

1 a a 1 a a

3 0 0= 0 3 0 = 3U

0 0 3

T *3 s sS = 3V I

* * *3 0 0 1 1 2 2S = 3V I + 3V I + 3V I

= Sum of symmtrical component powers

*0*

0 1 2 1*2

I= V + V + V I

I

Sequence Network for Power System Components

Impedance Load:

Balanced Star Grounded Load

a

c

b

ZY

In

ZY ZY

Ib

g

N Zn

Ia

Ic

Vag Vcg

Vbg

+

+

+---

ag Y a n nV = Z I + Z I

Y a n a b c= Z I + Z (I +I +I )

Y n a n b n c= (Z + Z )I + Z I + Z I

bg n a Y n b n cV = Z I + (Z + Z )I + Z I

cg n a n b Y n cV = Z I + Z I + (Z + Z )I

ag Y n n n a

bg n Y n n b

cg n n Y n c

V (Z + Z ) Z Z IV = Z (Z + Z ) Z IV Z Z (Z + Z ) I

p p pV = Z I s p sAV = Z AI

- 1s p sV = (A Z A)I s s sV = Z I

- 1s pZ = A Z AWhere,

Y n n n2

s n Y n n2

n n Y n

2

2

1 1 1 (Z + Z ) Z Z1Z = 1 a a Z (Z + Z ) Z3 1 a a Z Z (Z + Z )

1 1 1 1 a a

1 a a

Y n

y

y

(Z +3Z ) 0 0= 0 Z 0

0 0 Z

0 Y n 0

1 Y 1

2 Y 2

V (Z +3Z ) 0 0 IV = 0 Z 0 IV 0 0 Z I

0 Y n 0 0 0V = (Z +3Z )I = Z I

1 Y 1 1 1V = Z I = Z I

2 Y 2 2 2V = Z I = Z I

Zero-sequence network

ZYI0

3ZnV0

+

-

Z0 = ZY + 3Zn

For star ungrounded load Zn = infinity

Zero sequence network is open

Positive-sequence network

ZY

I1V1

+

-

Z1 = ZY

Negative-sequence network

ZY

I2V2

+

-

Z2 = Z1 = ZY

Balanced Delta Connected Load:

1

ZZ =3

2 1

ZZ = Z =3

I1

I2

Z3

Z0 =

I0Z

3

Z

Z

Z

a

b

c

General three-phase impedance

load Ib

g

IaIc

Vag VcgVbg

+

---

+

+

General three-phase impedance load

- 1s pZ = A Z A

0 01 02 aa ab ac

210 1 12 ab bb bc

220 21 2 ac bc cc

2

2

Z Z Z 1 1 1 Z Z Z1 Z Z Z = 1 a a Z Z Z 3Z Z Z 1 a a Z Z Z

1 1 1 1 a a

1 a a

0 aa bb cc ab ac bc1Z = (Z + Z + Z +2Z +2Z +2Z )3

1 2 aa bb cc ab ac bc1Z = Z = (Z + Z + Z - Z - Z - Z )3

2 201 20 aa bb cc ab ac bc

1Z = Z = (Z +a Z +aZ - aZ - a Z - Z )3

2 202 10 aa bb cc ab ac bc

1Z = Z = (Z +aZ +a Z - a Z - aZ - Z )3

2 212 aa bb cc ab ac bc

1Z = (Z +a Z +aZ +2aZ +2a Z +2Z )3

2 221 aa bb cc ab ac bc

1Z = (Z +aZ +a Z +2a Z +2aZ +2Z )3

aa bb ccZ = Z = Z

ab ac bcZ = Z = Zand

Conditions for a symmetrical load

01 10 02 20 12 21Z = Z = Z = Z = Z = Z = 0

0 aa abZ = Z +2Z 1 2 aa abZ = Z = Z - Z

Example: Three identical Y-connected resistors from a load bank with a three phase rating of 2300 V and 500 KVA. If each bank has applied voltages

|Vab| = 1800V |Vbc| = 2700V |Vca| = 2300V

find the line voltages and currents in per unit into the load. Assume that the neutral of the load is not connected to the neutral of the system and select a base of 2300 V,100 KVA.

Solution: On 2300 V, 100 KVA base line voltages in per unit are: abV = 1800 2300 = .7826

bcV = 2700 2300 = 1.174 caV = 2300 2300 = 1.0

0

ca

0 0ab bc

Let, V 180 be taken as referenceThen, V = .7826 81.39 and V = 1.174 - 41.23Symmetrical components of the line voltages are

(

ab1V = 1/3[.7826 81.39 + 1.174 120 - 41.23) + 1.0 (240 +180)]

2ab1 ab bc ca

1V = (V +aV +a V )3

= 1/3[.1171+j.7738+.2286+j1.152+.5+j.866]=1/3[.8454+j2.7918] = .2819+ j.9306 = 0.97236 073.147

(

0 0 0ab2

0 0

V = 1/3[.7826 81.39 + 1.174 240 - 41.23 ) + 1.0 (120 +180 )]= 1/3[0.11716+j0.7737-1.111-j0.378 + 0.5 - j0.866]=1/3[-0.49339 - j0.4703] = -0.1645 - j0.1567 =.2272 =.2272 0-136.37 0223.627

2ab2 ab bc ca

1V = (V +a V +aV )3

0 0 0ab0V =1/3[.7826 81.39 + 1.174 - 41.23 + 1.0 180 ] = 0

Voltages to neutral are given by

1 0 0 0anV = 0.97236 (73.147 - 30 ) = 0.97236 43.147

ab0 ab bc ca1V = (V + V + V )3

Vca

Vbc Vabn

1 0 0aI = (0.97236 43.147 )/.2 = 4.8618 43.147 p.u. 2 0 0aI = (0.2272 253.627 )/.2 = 1.136 253.627 p.u.

0aI = 0

2 0 0 0anV = 0.2272 (223.627 + 30 ) = 0.2272 253.6270anV = 0resistance value in p.u.=1.100/500=.2 p.u.

SEQUENCE NETWORKS

SEQUEMCE NETWORKSLesson Summary1. Introduction2. Sequence networks for Transmission

Lines3. Sequence network for Genetators4. Sequence network for Transformers5. Sequence network for Power System

Instructional Objective

On completion of this lesson a student should be able to:

A. Develop sequence network Tr. Lines

B. Develop sequence network for Generators

C. Develop sequence network for Transformers

D. Assemble sequence networks for small power systems

Sequence Networks of Three-Phase LinesZaa

Zbb

Zcc Zbc

ZabZca

ZnnZbn

Zan

Zcn

Ia

Ib

Ic

InVcn

Vcn

Van

Vcn

Vbn

Van

a

b

c

n

a

b

c

n

For a fully transposed line:

Zaa = Zbb = Zcc Zab = Zbc = ZcaZan = Zbn = Zcn

Ia Zab Ib Zab Ic Zan In Zaa+ - + - + -

In Zan Ia Zan Ib Zan Ic Znn+ - + - + -

a a

n n

Van Van

an aa a ab b ab c an n a'n'

nn n an c an b an a

V = Z I + Z I + Z I + Z I + V - (Z I + Z I + Z I + Z I )

an a'n' aa an a ab an b c

an nn n

V - V = (Z - Z )I +(Z - Z )(I +I )+(Z - Z )I

bn b'n' aa an b ab an a c

an nn n

V - V = (Z - Z )I +(Z - Z )(I +I )+(Z - Z )I

cn c'n' aa an c ab an a b

an nn n

V - V = (Z - Z )I +(Z - Z )(I +I )+(Z - Z )I

n a b cI = - (I +I +I )

an a'n' aa nn an a ab nn an b

ab nn an c

V - V = (Z + Z - 2Z )I +(Z + Z - 2Z )I+(Z + Z - 2Z )I

bn b'n' ab nn an a aa nn an b

ab nn an c

V - V = (Z + Z - 2Z )I +(Z + Z - 2Z )I+(Z + Z - 2Z )I

cn c'n' ab nn an a ab nn an b

aa nn an c

V - V = (Z + Z - 2Z )I +(Z + Z - 2Z )I+(Z + Z - 2Z )I

s aa nn anZ Z + Z - 2Z

m ab nn anZ Z + Z - 2Z

aa' an a'n' s m m a

bb' bn b'n' m s m b

cc' cn c'n' m m s c

V V - V Z Z Z IV = V - V = Z Z Z I V V - V Z Z Z I

aa' an a'n'V V - V

bb' bn b'n'V V - V

cc' cn c'n'V V - V 0

sy 1

2

Z 0 0Z = 0 Z 0

0 0 Z

- 1sy pZ = A Z A

0 s mZ = Z +2Z 1 2 s mZ = Z = Z - Z

0 0' 0 0V - V = Z I 1 1' 1 1V - V = Z I

2 2' 2 2V - V = Z I


I0

V0+

-

Z0 = Zaa + 2Zab

V0

+

-


I1

V1

+

-V1

+

-

Z1 = Zaa - Zab


I1

V2+

-V2

+

-

Z2 = Z1 = Zaa - Zab

Sequence networks of Synchronous Generator

a

c

b

Zn

Ic

EcIa

Ib

InEa

Eb

++

+

-- -

I1


+

-

V1Eg

Zg1

+-

g1 d

'' ''g g g1 d

Z jXFor calculating initial Sub - transient fault current E = E and Z jX

I2


+

-V2

Zg2

g2

'q d

g2

'' '' '' ''d q d

X X Z j

2For calculating initial Sub - transient fault current

Z jX as X X

I0


+

-

V03Zn

Zg0

g0 l

g1 g2 g0

Z jX Z Z > Z

Sequence networks of Synchronous Motor and Induction Motor

I1


+

-V1 Em1

Zm1 I1


+

-V1

Zm1

+

-

Synchronous motor Induction motor

I2


+

-V2

Zm2 I2


+

-V2

Zm2


I0


+

-V0 3Zn

Zm0


I0+

-V0 3Zn

Zm0


Per-Unit Sequence Models of Three-Phase Two-Winding Transformers

+ +

VL

-

VH

-

jXl 030

jXl

jXl

positive-sequence network

negative-sequence network

+VH1-

+VL1-

+VH2-

+VL2-

AB

C

H1

H2

H3

ZN

N a

b

c

X1

X2

X3

Zn

Schematic representation

Zero Sequence Network

AB

C

H1

H2

H3

ZN

N

ab

c

X1

X2

X3

Schematic representation

Transformer

+VH0

-

+VL0-


jXl+3ZN+3Zn

ZN Zn

Transformer

+VH0-

+VL0-


jXl+3ZN

Transformer

+

VH0 -

+VL0-


jXl

Transformer

+VH0

-

+VL0

-


jXl+3ZN

Transformer

+VH0

-

+VL0

-


jXl

Transformer

+VH0

-

+VL0-


jXl

The per-unit sequence network of the Y- transformer, shown in Figure (b), have thefollowing features.

1. The per-unit impedances do not depend onthe winding connections. That is, the per-unit impedances of a transformer that isconnected Y-Y. Y-, -Y. or - are thesame. However, the base voltages dodepend on the winding connections.

2. A phase shift is included in the per-unitpositive- and negativesequence networks.For the American standard, the positive-sequence voltages and currents on thehigh-voltage side of the Y- transformerlead the corresponding quantities on thelow-voltage side by 30. For negativesequence, the high-voltage quantities lagby 30.

3. Zero-sequence currents can flow in the Ywinding if there is a neutral connection, andcorresponding zero-sequence currents flowwithin the winding. However, no zero-sequence current enters or leaves the winding.


Example: For the system shown in figure assemble the sequence networks

Xg1 XT1 XL1 XT1 Xm1

Eg1 Em1


Xg2 XT2 XL2 XT2 Xm2


Xg0 XT0 XL0 XT0 Xm0


Thank You !

short circuit analysis lecture notes

Documents